(a) (i) An AM signal s(t) = A_c[1 + kₐ m(t)]cos(2πf_c t) is applied to the system shown in the figure. Show that the message signal m(t) can be obtained from the square-rooter output v₃(t): Assume that |kₐ m(t)|

Question

(a) (i) An AM signal s(t) = A_c[1 + kₐ m(t)]cos(2πf_c t) is applied to the system shown in the figure. Show that the message signal m(t) can be obtained from the square-rooter output v₃(t): Assume that |kₐ m(t)| < 1 for all t, the message signal m(t) is limited to the interval −ω ≤ f ≤ ω, and the carrier frequency f_c > 2ω. (10 marks)

(ii) A narrow band FM signal is approximately given as s(t) ≈ A_c cos(2π f_c t) - β A_c sin(2π f_c t)sin(2π fₘ t) Determine the envelope of this modulated signal. Also determine the ratio of the maximum to the minimum value of this envelope. Plot this ratio versus β, with β restricted to the interval 0 ≤ β ≤ 0·4. Also determine the average power of the narrow band FM signal, expressed as a percentage of the average power of the unmodulated carrier wave. (10 marks)

(b) (i) Explain why PWM inverters are preferred over square wave inverters. Further, draw the harmonic spectrum to highlight the differences in unipolar and bipolar PWM techniques. (10 marks)

(ii) A single-phase, full-bridge inverter has DC-link voltage V_DC = 400 V, and the fundamental frequency of 50 Hz. Find the r.m.s. value of the voltages of the fundamental and next two prominent harmonics for the following cases: (1) Square wave mode (2) Voltage cancellation mode with α = 20° (10 marks)

(c) A 50 hp, 440 V, 50 Hz, star-connected, three-phase induction motor has a starting torque of 75% and maximum torque of 250% of the full-load torque. Find the following: (i) Slip at which maximum torque occurs (ii) Slip at full-load torque (10 marks)

UPSC Answer Check · Accepted Answer

(a) (i)
Use cos²x=(1+cos2x)/2. Then
s(t)² = A_c²[1+k_a m(t)]² cos²(2πf_c t)
= (A_c²/2)[1+k_a m(t)]² + (A_c²/2)[1+k_a m(t)]² cos(4πf_c t).
The squarer output v_1(t) contains a DC term, a message term proportional to k_a m(t), a message-squared term proportional to k_a² m(t)², and the same terms at 2f_c. Because m(t) is limited to |f|≤ω, [1+k_a m(t)]² is limited to |f|≤2ω. The second term is centred at 2f_c. Since f_c>2ω, a low-pass filter with cutoff f_L satisfying 2ω < f_L < 2f_c − 2ω (the lower edge of the band around 2f_c) removes the 2f_c term and gives
v_2(t)=(A_c²/2)[1+k_a m(t)]².
Since |k_a m(t)|<1, the bracket is positive, so the square-rooter is exact and
v_3(t)=√(v_2(t))=(A_c/√2)[1+k_a m(t)].
Thus m(t)=(√2 v_3(t)/A_c - 1)/k_a. This is a DC offset and scale. A_c is in V, k_a is dimensionless, f_c and ω are in Hz, v_2 is in V², and v_3 is in V.

(a) (ii)
Use the quadrature-envelope method. Write
s(t)=A_c cosω_c t - βA_c sinω_m t sinω_c t = X cosω_c t + Y sinω_c t,
with X=A_c and Y=-βA_c sinω_m t. For f_c≫f_m, the envelope is
E(t)=√(X²+Y²)=A_c√(1+β² sin²ω_m t).
Thus E_max=A_c√(1+β²), E_min=A_c, and
R(β)=E_max/E_min=√(1+β²).
For 0≤β≤0.4, plot R versus β as y=√(1+β²); it is monotonic because dR/dβ=β/√(1+β²)≥0, and it passes through (0,1), (0.2,1.0198), and (0.4,1.0770).

(b) (ii)
For the square wave, v(θ)=V_DC for 0<θ<π and -V_DC for π<θ<2π. The sine coefficient is
b_n=(2/π)∫_0^π V_DC sin nθ dθ=(2V_DC/π n)(1-cos nπ).
Thus b_n=0 for even n and b_n=4V_DC/(π n) for odd n. The alternating signs do not affect rms values. The rms value is
V_n,rms=|b_n|/√2=(4V_DC)/(π n√2).
The full-wave rms is V_DC, and the fundamental is 0.9 V_DC; the remaining power is in harmonics. With V_DC=400 V and f_1=50 Hz, taking the next two prominent harmonic orders as 3rd and 5th:
Fundamental, 50 Hz: V_1,rms=1600/(π√2)=360.2 V.
Third, 150 Hz: V_3,rms=120.1 V.
Fifth, 250 Hz: V_5,rms=72.0 V.
For the standard voltage-cancellation waveform, α is the angle by which voltage is cancelled at each end of a half-cycle; α=0 reduces to the square wave. The output is zero for 0<θ<α and π-α<θ<π, and +V_DC for α<θ<π-α; the negative half-cycle is symmetric. For odd n,
b_n=(2/π)∫_α^(π-α) V_DC sin nθ dθ
=(2V_DC/π n)(-cos n(π-α)+cos nα).
For odd n, cos(nπ-nα)=-cos nα, so
b_n=(4V_DC)/(π n)cos nα.
Hence
V_n,rms=(4V_DC)/(π n√2)|cos nα|.
With α=20°:
Fundamental: V_1,rms=360.2 cos20°=338.5 V.
Third: V_3,rms=120.1 cos60°=60.0 V.
Fifth: V_5,rms=72.0 |cos100°|=12.5 V.
The 5th is small because cos100° is small, showing the cancellation effect. The 7th would be 39.4 V, but the next two harmonic orders are 3rd and 5th.

(c) (i)
Use T(s)∝R_2 s/(R_2²+s²X_2²). Let s_m=R_2/X_2. Then T(s)∝s_m s/(s_m²+s²). Differentiating with respect to s and setting dT/ds=0 gives s=s_m, so T_max occurs at s_m. Also T_max∝1/2 and T_st∝s_m/(1+s_m²). Given T_st=0.75T_fl and T_max=2.5T_fl,
T_max/T_st=10/3=(1+s_m²)/(2s_m).
Thus 3s_m²-20s_m+3=0, so s_m=(10±√91)/3. The torque curve has two roots for a torque below T_max; the normal motoring maximum is the low-slip root below 1:
s_m=(10-√91)/3=0.1535, i.e. 15.35%.

(c) (ii)
Let full-load slip be s_f. Then
T_max/T_fl=(s_m²+s_f²)/(2s_m s_f)=2.5.
Thus s_f²-5s_m s_f+s_m²=0, so s_f=s_m(5±√21)/2. The larger root lies above s_m and is not the normal full-load slip. The full-load point is on the low-slip side, so
s_f=s_m(5-√21)/2=0.0320, i.e. 3.20%.
Check: T_fl/T_max=0.4 and T_st/T_fl=0.75, consistent with the data. The hp, voltage, and frequency data are not needed because the torque ratios fix the slips.

(a) (i) An AM signal s(t) = A_c[1 + kₐ m(t)]cos(2πf_c t) is applied to the system shown in the figure. Show that the message…

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